Problem: Let $y=\dfrac{\ln(x)}{\sin(x)}$. Find $\dfrac{dy}{dx}$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{1-x\cos(x)}{x\sin(x)}$ (Choice B) B $\dfrac{\cos(x)+x\sin(x)\ln(x)}{x\sin(x)}$ (Choice C) C $\dfrac{1}{x\sin(x)}$ (Choice D) D $\dfrac{\sin(x)-x\ln(x)\cos(x)}{x\sin^2(x)}$
$\dfrac{\ln(x)}{\sin(x)}$ is the quotient of two, more basic, expressions: $\ln(x)$ and $\sin(x)$. Therefore, $\dfrac{dy}{dx}$ can be found using the quotient rule : $\begin{aligned} \dfrac{d}{dx}\left[\dfrac{u(x)}{v(x)}\right]&=\dfrac{\dfrac{d}{dx}[u(x)]v(x)-u(x)\dfrac{d}{dx}[v(x)]}{[v(x)]^2} \\\\ &=\dfrac{u'(x)v(x)-u(x)v'(x)}{[v(x)]^2} \end{aligned}$ Let's differentiate! $\begin{aligned} &\phantom{=}\dfrac{dy}{dx} \\\\ &=\dfrac{d}{dx}\left(\dfrac{\ln(x)}{\sin(x)}\right) \\\\ &=\dfrac{\dfrac{d}{dx}(\ln(x))\sin(x)-\ln(x)\dfrac{d}{dx}(\sin(x))}{(\sin(x))^2}&&\gray{\text{The quotient rule}} \\\\ &=\dfrac{\dfrac1x\cdot \sin(x)-\ln(x)\cdot \cos(x)}{(\sin(x))^2}&&\gray{\text{Differentiate }\ln(x)\text{ and }\sin(x)} \\\\ &=\dfrac{\sin(x)-x\ln(x)\cos(x)}{x\sin^2(x)}&&\gray{\text{Simplify}} \end{aligned}$ In conclusion, $\dfrac{dy}{dx}=\dfrac{\sin(x)-x\ln(x)\cos(x)}{x\sin^2(x)}$